hi guys,i have a 14.4v drill with no charger so i thought i would try something.
i have a 24v 0.75a wallwart and i cut the tip off of it,to see if it would charge my battery. i only left it hook up for a few minutes and it charged my battery somewhat.when i unplugged the wallwart it was really hot so i new not to try that no more. my question is why did the wallwart get so hot,i dont understand how im sending voltage to the battery makes the wallwart get so hot. thanks will41

 

Maybe the wall-wart is AC instead of DC. You didn't say which it is.
Maybe you connected the polarity backwards.

 

it says ac adapter on it

input ac 100-120v-o.4a
output 24v 0.75a in between the 24v and 0.75a is a straight line underneath it is 3 short lines example ---

 

ok i look up the synbol it is dc i was looking it to say dc but it just as a symbol.how would i go about building a charger for my drill.
like i said before just connecting the wires to the battery from the wallwart,causes the the wallwart to get hot.so i know that isnt the right way of doing this.i am trying to learn hereso any help would be appreciated thank you will41

 

A battery charger limits its current to match the spec's of the battery it charges.
Your wall-wart did not limit the current so it gets too hot.

A battery charger is supposed to detect when the battery is fully charged then switch to a low current trickle charge or turn off.
Your wall-wart did neither.

 

thanks alot
so is there a way i can use this wallwart to charge my battery pack
without it overheating

 

Attached is a schematic for a simplified battery charger using an LM317 voltage regulator.

Rlim should actually be an automotive taillight bulb like an 1157, both filaments connected in parallel. The light bulb will limit the maximum current when the filaments are hot. As the current through the bulb decreases when the battery gets more charged, the bulb's resistance decreases.

LM317 regulators are available at any electronics store. You will need to attach a large heat sink to the regulator.

Rs causes the LM317 current output to be fairly constant (high) until the battery is nearly charged.
R1 sets the Vref current between OUT and ADJ (the LM317 terminal marked COM). R2 sets the output voltage. This must be adjusted to 14.4v prior to connecting a battery, or the battery will be over- or under- charged.

R3 ensures that the regulator's minimum 10mA current requirement is met once the battery is charged. This keeps the regulator's output stable.

C1 is necessary to prevent oscillation of the regulator in case it is operated without a battery connected, or if the wires to the battery are more than a few inches long.

 

in the schematic where it says Rs .2 is that a resistor
im thrown at that point everything i pretty much get thank you

 

Look up using the LM317 as a current regulator, that is for current limiting.

 

Yes, that is an 0.2 Ohm resistor. It sets the source impedance from the LM317. It keeps the charge current pretty constant until the battery voltage comes up near the set voltage.